Question

At the Foremost State Bank the average savings account balance in 2012 was $1900. A random sample of 45 savings account balanes for 2013 yielded a mean of $1000 with a standard deviation of $225. At the α = 0.01 significance level test the claim that the mean savings account balance in 2013 is different from the mean savings account balance in 2012.

Answer 1

Answer:

The calculated  |t| = |-26.517|=26.517 > 2.326 at 0.01 level of significance

Null hypothesis is rejected at 0.01 level of significance

There is difference between mean savings account balance in 2013 is different from the mean savings account balance in 2012

Step-by-step explanation:

Step(i):-

Mean of the Population = 1900

Given random sample size 'n' =  45

Mean of the sample x⁻ = 1000

Standard deviation of the sample = 225

Null Hypothesis:-

There is no difference between mean savings account balance in 2013 is different from the mean savings account balance in 2012.

Alternative Hypothesis :-

There is difference between mean savings account balance in 2013 is different from the mean savings account balance in 2012

Step(ii):-

Test statistic

        $t = \frac{x^{-}-mean }{\frac{S.D}{\sqrt{n} } }$

        $t = \frac{1000-1900 }{\frac{225}{\sqrt{45} } }$

       t = -900 /33.54 = -26.517

      |t| = |-26.517|=26.517

Degrees of freedom

                       γ = n-1 = 45-1 =44

t₍₀.₀₁ , ₄₄₎ = 2.326

The calculated  |t| = |-26.517|=26.517 > 2.326 at 0.01 level of significance

Null hypothesis is rejected at 0.01 level of significance

There is difference between mean savings account balance in 2013 is different from the mean savings account balance in 2012